## Precalculus (6th Edition) Blitzer

Consider the system of equations, \begin{align} & x+y+z+w=-1 \\ & -x+4y+z-w=0 \\ & x-2y+z-2w=11 \\ & -x-2y+z+2w=-3 \end{align} First we will write the augmented matrix for the given system of equations: Augmented matrix is: $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ -1 & 4 & 1 & -1 \\ 1 & -2 & 1 & -2 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ 0 \\ 11 \\ -3 \\ \end{matrix} \right]$ Now, using the row operation, we will reduce the matrix to row echelon form. ${{R}_{2}}\to {{R}_{2}}+{{R}_{1}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 1 & -2 & 1 & -2 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 11 \\ -3 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 0 & -3 & 0 & -3 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 12 \\ -3 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{1}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 0 & -3 & 0 & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 12 \\ -4 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{5}{{R}_{2}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & -3 & 0 & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ 12 \\ -4 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+3{{R}_{2}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & \frac{6}{5} & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{57}{5} \\ -4 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{2}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & \frac{6}{5} & -3 \\ 0 & 0 & \frac{12}{5} & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{57}{5} \\ \frac{-21}{5} \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{5}{6}{{R}_{3}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & \frac{12}{5} & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ \frac{-21}{5} \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}-\frac{12}{5}{{R}_{3}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 & 9 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ -27 \\ \end{matrix} \right]$ ${{R}_{4}}\to \frac{1}{9}{{R}_{4}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ -3 \\ \end{matrix} \right]$ Now, the reduced system of equations is given by: $x+y+z+w=-1$ …… (I) $y+\frac{2}{5}z=\frac{-1}{5}$ …… (II) $z-\frac{5}{2}w=\frac{19}{2}$ …… (III) $w=-3$ …… (IV) Substitute the value of $w$ in equation (III), Thus, \begin{align} & z-\frac{5}{2}\times -3=\frac{19}{2} \\ & z+\frac{15}{2}=\frac{19}{2} \\ & z=\frac{19}{2}-\frac{15}{2} \\ & z=\frac{4}{2} \end{align} After solving it further we get, Then, $z=2$ Now substitute the values of $z$ in equation (II) to get, \begin{align} & y+\frac{2}{5}\times 2=\frac{-1}{5} \\ & y+\frac{4}{5}=\frac{-1}{5} \\ & y=\frac{-1}{5}-\frac{4}{5} \\ & y=\frac{-5}{5} \\ \end{align} After solving it further we get, Then, $y=-1$ Now substitute the values of $y,z,w$ in equation (I) to find the value of x, Thus, \begin{align} & x-1+2-3=-1 \\ & x-2=-1 \\ & x=1 \end{align} Therefore, $x=1,y=-1,z=2,w=-3$ Hence, the provided statement makes sense.